(Breaking the affine cipher, 15pt). You\'ve overheard the ciphertext \'FZUNWBF\'

Question

(Breaking the affine cipher, 15pt). You've overheard the ciphertext 'FZUNWBF'. By bribing somebody you were able to find out that the last three letters of the corresponding ciphertext are 'nym'.

a) [7pt] Find the encryption key, i.e. determine a (alpha) and b (beta) so that ax+b (mod 26) encrypts '****nym' as 'FZUNWBF'.

b) [4pt] Determine the corresponding decryption key for the encryption key you found in a).

c) [4pt] Apply the decryption key from b) to find the original plaintext of  'FZUNWBF'.

Explanation / Answer

(a) n is encrypted as W, y as B and m as F. So,

n= 13 W= 22

y= 24 B= 1

m= 12 F= 5

We have c= ax+b (mod 26)

22= 13a+b (mod 26).........(1)

Also,

1= 24a+b (mod 26)...........(2)

Using (1) and (2), we get

21= -11a mod(26)

For a= 17 above equation satisfied (By hit and trial)

Therefore

22 = 13*17 + b (mod 26)

22= 221 + b (mod 26)

Therefore b = 9

We can verify these for m and F

12*17+ 9 (mod 26) = 213 mod 26 = 5

and 5 represents F.

Therefore keys are

a= 17, b = 9

(b) 17x + 9 (mod 26) = c

x= (17)-1(c- 9) (mod 26)

The inverse of 17 is 23 as (17*23) mod 26 = 1

x= (23c - 23*9) (mod 26)

x= (23c -207) (mod 26)

x= 23c +1 (mod 26)

So decryption can be done by above method.

for F=5

23*5 + 1 = 116 (mod 26) = 11 = m

for Z= 25

23*25 +1= 576 mod 2

= 4 = e

for U = 20

23*20+1= 461 mod 26 =19

= t

for N= 13

23*13+1 = 300 mod 26 = 14

=o

for W = 22

23*22+1 = 507 mod 26

= 13 = n

for B,

y

for F,

m

So, original plaintext is "metonym".

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